Question

More Linear Algebra. Latexing it up below. Why doesn't OpenStudy have the option to have Latex it in the question portion?

Answer 1

How does one find the eigenvalues through inspection or by quick algebra?
Say given matrix A = \[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix}\]

How does one find the eigenvalues of this?

Answer 2

Well one quick way is that the trace of a matrix is the sum of its eigenvalues and the determinant of a matrix is the product of its eigenvalues.

It's really helpful when dealing with 2x2 matrices, and kinda with 3x3, but in general I don't know. I'm sure there's a trick for this one since the columns are all linearly dependent.

Answer 3

The hints were:

Note that the algebraic multiplicity agrees with the geometric multiplicity.
(Why?) Hint: What is the kernel of A?

The answer is that
ker(A) is four-dimensional, so that the eigenvalue 0 has multiplicity 4, and the remaining eigenvalue is
tr(A) = 5.

Answer 4

I just don't understand why they knew that the eigenvalue was 0 off the bat

Answer 5

In this case, because the entire matrix is linearly dependent, you know that the matrix's characteristic will have a 0 solution. Think back to what an eigenvalue really represents. It's a number we could multiply by an eigenvector instead of the matrix. Since the columns are linearly dependent, we can always find a way to combine all the columns to get 0, in 4 different ways without the vectors being 0 to begin with.

\[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -1 \end{bmatrix}\]

Here's one of 4 ways, here's another way:

\[\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \\0 \end{bmatrix} = 0 \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \\ 0 \end{bmatrix}\]

So maybe it becomes more transparent now how these are eigenvalues satisfying \[A x= \lambda x\]

Really what it comes down to is identifying the linearly dependent columns, that's what's going to give you the power to easily say this kinda thing.

Answer 6

So, using this to check a hypothesis: http://www.mathportal.org/calculators/matrices-calculators/matrix-calculator.php


It seems as if per every redundant column, we get a 0 eigenvector. So in this case, given 4 redundant columns. We get a 0 with an almu of 4 (# of redundant columns). Is this true?